Calculus Problem: Showing Partial Derivatives Equal Each Other

[Mathman23]

Hi

Given a function

z = f(x,y), where x = r * cos(\phi) and y = r * sin (\phi)

First I show that

\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} cos(\phi) + \frac{\partial z}{\partial y} sin (\phi)

and

\frac{\partial z}{\partial \phi} = - \frac{\partial z}{\partial x} r \cdot sin(\phi) + \frac{\partial z}{\partial y} r \cdot sin(\phi)

Finally I need to show that

(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi}) ^2

How do I approach this part of the problem?

Sincerley

Fred

 

[arildno]

Hmm..square both previous equations and add them together, mayhap?

 

[Mathman23]

This is my solution for (b) please look at them at see if I made a mistake:

solving (b)


(\frac{\partial z}{\partial x}^2) ^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi})^2 = (\frac{\partial z}{\partial x})^2 \cdot cos ^2 (\phi) + (\frac{\partial z}{\partial y})^2 \cdot sin(\phi) + 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} sin(\phi) \cdot cos(\phi) + (\frac{\partial z}{\partial x}) ^2 \cdot \frac{r^2 \cdot sin^2 (\phi)}{r^2} + (\frac{\partial z}{\partial y})^2 \cdot \frac{r^2 \cdot cos^2 (\phi)}{r^2} - 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} \cdot \frac{r^2 sin(\phi) \cdot cos(\phi)}{r^2} =

= (\frac{\partial z}{\partial x})^2 \cdot (cos^2(\phi) + sin^2 (\phi) \cdot (\frac{\partial z}{\partial y})^2 sin^2 (\phi) = (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2


Sincerely Yours
Fred